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A002162
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Decimal expansion of the natural logarithm of 2.
(Formerly M4074 N1689)
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106
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6, 9, 3, 1, 4, 7, 1, 8, 0, 5, 5, 9, 9, 4, 5, 3, 0, 9, 4, 1, 7, 2, 3, 2, 1, 2, 1, 4, 5, 8, 1, 7, 6, 5, 6, 8, 0, 7, 5, 5, 0, 0, 1, 3, 4, 3, 6, 0, 2, 5, 5, 2, 5, 4, 1, 2, 0, 6, 8, 0, 0, 0, 9, 4, 9, 3, 3, 9, 3, 6, 2, 1, 9, 6, 9, 6, 9, 4, 7, 1, 5, 6, 0, 5, 8, 6, 3, 3, 2, 6, 9, 9, 6, 4, 1, 8, 6, 8, 7
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OFFSET
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0,1
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COMMENTS
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Newton calculated the first 16 terms of this sequence.
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REFERENCES
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S. R. Finch, Mathematical Constants, Cambridge, 2003, Sections 1.3.3 and 6.2.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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Harry J. Smith, Table of n, a(n) for n = 0..20000
D. H. Bailey and J. M. Borwein, Experimental Mathematics: Examples, Methods and Implications, Notices of the AMS, May 2005, Volume 52, Issue 5.
P. Bala, New series for old functions
Paul Cooijmans, Odds.
X. Gourdon and P. Sebah, The logarithm constant:log(2)
I. Newton, The method of fluxions and infinite series; with its application to the geometry of curve-lines, 1736; see p. 96.
Simon Plouffe, log(2), natural logarithm of 2 to 2000 places
S. Ramanujan, Question 260, J. Ind. Math. Soc.
D. W. Sweeney, On the computation of Euler's constant, Math. Comp., 17 (1963), 170-178.
Horace S. Uhler, Recalculation and extension of the modulus and of the logarithms of 2, 3, 5, 7 and 17, Proc. Nat. Acad. Sci. U. S. A. 26, (1940). 205-212.
Eric Weisstein's World of Mathematics, Natural Logarithm of 2, Masser-Gramain Constant, Logarithmic Integral
Wikipedia, Natural logarithm of 2.
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FORMULA
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log(2) = Sum_{k>=1} 1/(k*2^k) = Sum_{j>=1} (-1)^(j+1)/j.
log(2) = Integral_{t=0..1} dt/(1+t).
log(2) = 2/3 * (1 + sum(k>=1, 2/[(4k)^3-4k] )) (Ramanujan).
log(2) = 4*sum_{k>=0} [3-2*sqrt(2)]^(2k+1)/(2k+1) (Y. Luke). - R. J. Mathar, Jul 13 2006
log(2) = 1-(1/2)*sum_{k>=1} 1/(k*(2k+1)). - Jaume Oliver Lafont, Jan 06 2009, Jan 08 2009
log(2) = 4*sum_{k>=0} 1/((4k+1)(4k+2)(4k+3)). - Jaume Oliver Lafont, Jan 08 2009
log(2) = 7/12+24*sum_{k>=1} 1/(A052787(k+4)*A000079(k)). - R. J. Mathar, Jan 23 2009
From Alexander R. Povolotsky, Jul 04 2009: (Start)
log(2) = 1/4*(3 - sum(n>=1, 1/(n*(n+1)*(2*n+1)) )).
log(2) = (230166911/9240-sum(k>=1, (1/2)^k* (11/k+10/(k+1)+9/(k+2)+8/(k+3)+7/(k+4)+6/(k+5)-6/(k+7)-7/(k+8)-8/(k+9) -9/(k+10)-10/(k+11)) ))/35917. (End)
log(2) = A052882/A000670. - Mats Granvik, Aug 10 2009
From log(1-x-x^2) at x=1/2, log(2)=(1/2)*Sum_{k>=1}L(k)/(k*2^k), where L(n) is the n-th Lucas number (A000032). - Jaume Oliver Lafont, Oct 24 2009
log(2) = Sum_{k>=1} 1/(cos(k*Pi/3)*k*2^k) (Cf. A176900). - Jaume Oliver Lafont, Apr 29 2010
log(2) = (sum(n>=1, 1/(n^2*(n+1)^2*(2*n+1)) ) +11)/16. - Alexander R. Povolotsky, Jan 13 2011
log(2) = sum(n>=1, (2*n+1)/(sum(k=1..n, k^2 )^2)) +396)/576. - Alexander R. Povolotsky, Jan 14 2011
log(2) = 105*(sum(n>=1, 1/(2*n*(2*n+1)*(2*n+3)*(2*n+5)*(2*n+7)) ) - 319/44100) log(2) = (319/420 - 3/2*sum(n>=1, 1/(6*n^2+39*n+63) )). - Alexander R. Povolotsky, Dec 16 2008
log(2) = sum(k>=1, A191907(2,k)/k ). - Mats Granvik, Jun 19 2011
log(2) = Integral_{x=0..oo} 1/(1 + e^x) dx. - Jean-François Alcover, Mar 21 2013
log(2) = limit of zeta(s)*(1-1/2^(s-1)) as s -> 1. - Mats Granvik, Jun 18 2013
From Peter Bala, Dec 10 2013: (Start)
log(2) = 2*sum {n = 1..inf} 1/( n*A008288(n-1,n-1)*A008288(n,n) ), a result due to Burnside.
log(2) = 1/3*sum {n >= 0} (5*n+4)/( (3*n+1)*(3*n+2)*C(3*n,n) )*(1/2)^n = 1/12*sum {n >= 0} (28*n+17)/( (3*n+1)*(3*n+2)*C(3*n,n) )*(-1/4)^n.
log(2) = 3/16*sum {n >= 0} (14*n+11)/( (4*n+1)*(4*n+3)*C(4*n,2*n) )*(1/4)^n = 1/12*sum {n >= 0} (34*n+25)/( (4*n+1)*(4*n+3)*C(4*n,2*n) )*(-1/18)^n. For more series of this type see the Bala link.
See A142979 for series acceleration formulas for log(2) obtained from the Mercator series log(2) = sum {n >= 1} (-1)^(n+1)/n. See A142992 for series for log(2) related to the root lattice C_n. (End)
log(2) = sum(k=2^n..2^(n+1)-1, 1/k) as n -> Infinity. - Richard R. Forberg, Aug 16 2014
From Peter Bala, Feb 03: (Start)
log(2) = 2/3*Sum {k >= 0} 1/((2*k + 1)*9^k).
Define a pair of integer sequences A(n) = 9^n*(2*n + 1)!/n! and B(n) = A(n)*sum {k = 0..n} 1/((2*k + 1)*9^k). Both satisfy the same second order recurrence equation u(n) = (40*n + 16)*u(n-1) - 36*(2*n - 1)^2*u(n-2). From this observation we obtain the continued fraction expansion log(2) = 2/3*(1 + 2/(54 - 36*3^2/(96 - 36*5^2/(136 - ... - 36*(2*n - 1)^2/((40*n + 16) - ... ))))). Cf. A002391, A073000 and A105531 for similar expansions. (End)
log(2) = Sum_{n>=1} (Zeta(2*n)-1)/n. - Vaclav Kotesovec, Dec 11 2015
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EXAMPLE
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0.693147180559945309417232121458176568075500134360255254120680009493393...
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MATHEMATICA
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RealDigits[N[Log[2], 200]][[1]] (* Vladimir Joseph Stephan Orlovsky, Feb 21 2011 *)
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PROG
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(PARI) { default(realprecision, 20080); x=10*log(2); for (n=0, 20000, d=floor(x); x=(x-d)*10; write("b002162.txt", n, " ", d)); } /* Harry J. Smith, Apr 21 2009 */
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CROSSREFS
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Cf. A016730 Continued fraction. A008288, A142979, A142992.
Sequence in context: A129938 A022698 A013707 * A257945 A271526 A072365
Adjacent sequences: A002159 A002160 A002161 * A002163 A002164 A002165
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KEYWORD
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cons,nonn
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AUTHOR
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N. J. A. Sloane, Apr 30 1991
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STATUS
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approved
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